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Tambaya 1 Rahoto
A typical chemical reaction will be spontaneous if
Bayanin Amsa
In thermodynamics, a chemical reaction is considered spontaneous when it occurs naturally under a given set of conditions without needing to be driven by an external force. The spontaneity of a reaction is best determined by the Gibbs Free Energy change, denoted as ΔG.
The criteria for spontaneity is as follows:
Now, let's relate this to the given options:
Thus, a chemical reaction is spontaneous when the Gibbs Free Energy change (ΔG) is negative.
Tambaya 2 Rahoto
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Bayanin Amsa
The question is asking about the pH of a 0.001 mol dm−3 solution of H2SO4 (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.
Step 1: Dissociation of H2SO4
Sulfuric acid, H2SO4, is a strong acid and dissociates completely in water in two steps:
1. The first dissociation: H2SO4 → H+ + HSO4-
2. The second dissociation: HSO4- → H+ + SO42-
For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H+ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.
Step 2: Calculate the H+ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H2SO4, we get 2 moles of H+. Therefore, for a 0.001 mol dm−3 solution of H2SO4, the concentration of H+ ions will be:
2 x 0.001 = 0.002 mol dm−3
Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H+]
Substitute the H+ concentration:
pH = -log(0.002)
We know that log(10-2) = -2 and log(2) = 0.3 (as provided), so:
pH = -(log(2) + log(10-3))
pH = -(0.3 - 3)
pH = 3 - 0.3
pH = 2.7
Therefore, the pH of the 0.001 mol dm−3 H2SO4 solution is 2.7.
Tambaya 3 Rahoto
If 11.0g of a gas occupies 5.6 dm3 at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm3 ).
Bayanin Amsa
The problem requires calculating the **vapor density** of the gas. Vapor density is defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen, where the hydrogen standard is 2 g/mol (as the molecular weight of hydrogen gas, H₂, is 2).
Here's a step-by-step explanation:
The calculated vapor density of the gas is 22.
Tambaya 4 Rahoto
The constituents of Alnico are Aluminium, Nickel and
Bayanin Amsa
Alnico is a type of alloy that is known for its strong magnetic properties. The name "Alnico" comes from the elements it is primarily composed of: Aluminum (Al), Nickel (Ni), and Cobalt (Co). These elements are combined to form an alloy that retains its magnetism well and can operate at high temperatures, making it ideal for applications like electric motors, sensors, and various electronic devices.
While there are different variations of Alnico, the presence of Cobalt (Co) is essential for enhancing the magnetic properties of the alloy. The other elements listed, such as Magnesium (Mg), Manganese (Mn), and Copper (Cu), are not typical core constituents of Alnico. Although trace amounts of other elements like copper may sometimes be included in specific formulations, the primary and most significant component responsible for Alnico's powerful magnetic characteristics is Cobalt (Co).
Tambaya 5 Rahoto
Bayanin Amsa
In the Contact Process, the catalyst used for the conversion of sulphur(IV) oxide (SO2) to sulphur(VI) oxide (SO3) is vanadium(V) oxide, also chemically represented as V2O5. This catalyst is preferred because it is more cost-effective and significantly more durable under reaction conditions than other catalysts such as platinum. Moreover, while platinum is also an effective catalyst, it is prone to poisoning by impurities that may be present in the reaction mixture. Vanadium(V) oxide, on the other hand, offers a better balance of efficiency, cost, and durability, making it the catalyst of choice in industrial applications of the Contact Process.
Tambaya 6 Rahoto
The amount of Faraday required to discharge 4.5 moles of Al3+ is
Bayanin Amsa
To determine the amount of Faraday required to discharge 4.5 moles of Al3+ ions, it is essential to understand Faraday's laws of electrolysis and the concept of moles in chemistry.
When discharging Al3+ ions to form aluminum metal (Al), the reduction half-reaction involved is:
Al3+ + 3e- → Al
From this equation, it can be seen that 3 moles of electrons (e-) are required to discharge 1 mole of Al3+ ions to form 1 mole of aluminum metal.
A Faraday is the amount of electric charge carried by one mole of electrons. Therefore, 1 Faraday corresponds to the charge needed to discharge 1 mole of electrons.
Now, to discharge 4.5 moles of Al3+, we need:
4.5 moles of Al3+ × 3 moles of electrons (e-)/mole of Al3+ = 13.5 moles of electrons
Since each Faraday discharges 1 mole of electrons, 13.5 moles of electrons correspond to 13.5 Faradays of charge.
Hence, the amount of Faraday required to discharge 4.5 moles of Al3+ ions is 13.5 Faradays.
Tambaya 7 Rahoto
An example of an amphoteric oxide is
Bayanin Amsa
An example of an amphoteric oxide is Al2O3 (aluminum oxide).
Amphoteric oxides are special because they can act as both acidic and basic oxides. This means they can react with both acids and bases to form salts and water, showcasing their dual behavior.
Here is how it works:
In contrast, oxides like CuO (copper(II) oxide) are basic oxides, and K2O (potassium oxide) is a basic oxide as well. They don't exhibit both acidic and basic properties.
Therefore, the amphoteric nature of Al2O3 is what distinguishes it from common oxides that are strictly acidic or basic. This property is crucial in various chemical processes and applications.
Tambaya 8 Rahoto
Hydrochloric acid is not suitable in the preparation of ethanoic acid because it
Bayanin Amsa
Hydrochloric acid is not suitable for preparing ethanoic acid because it is too volatile.Being too volatile, means it has a low boiling point and is easily evaporated. Thus, HCl is not suitable because it cannot carry out the oxidation process required to convert alcohols into acids like ethanoic acid.
Ethanoic acid, also known as acetic acid, is a weak acid that doesn't fully dissociate in water, while hydrochloric acid is a strong acid that dissociates almost completely.
Tambaya 9 Rahoto
Water gas obtained from the gasification of coke is made up of
Bayanin Amsa
The gasification of coke to produce water gas involves reacting coke, which is primarily composed of carbon, with steam. The main chemical reaction that occurs is:
C (s) + H2O (g) → CO (g) + H2 (g)
From this reaction, the main constituents of water gas are hydrogen (H2) and carbon monoxide (CO), also known as carbon(II) oxide. Therefore, water gas obtained from the gasification of coke is made up of hydrogen and carbon(II) oxide.
Tambaya 10 Rahoto
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Bayanin Amsa
The reaction between hydrogen and chlorine to produce hydrogen chloride gas is explosive in sunlight. This is because sunlight contains a broad range of electromagnetic radiation, including ultraviolet (UV) light, which is energetic enough to initiate the reaction.
Here is a simplified explanation:
In contrast, other forms of light like diffused light, infrared light, and Raman light do not provide enough energy to initiate this explosive reaction because they lack the necessary UV component found in sunlight.
Tambaya 11 Rahoto
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Bayanin Amsa
To answer this question, let's break it down step by step:
Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.
Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).
If we symbolize the number of protons by the atomic number (Z), we can say:
1. **Protons = Electrons** in a neutral atom.
2. **Mass Number (A) = Protons + Neutrons**.
Given that the mass number is 31, we have the equation:
A = Protons + Neutrons = 31.
Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:
1. **Protons = 15**.
2. **Electrons = 15** (because it's a neutral atom).
3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.
So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.
Tambaya 12 Rahoto
Alkanoates are naturally found in
Bayanin Amsa
Alkanoates, also known as fatty acid esters, are primarily found in lipids. Lipids are a broad group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins (such as vitamins A, D, E, and K), and others. One of the main components of lipids is fatty acids and their derivatives, such as alkanoates.
To be more specific, alkanoates can be found in the form of triglycerides, which are the main constituents of body fat in humans and animals, as well as vegetable fat. Triglycerides are composed of glycerol bound to three fatty acids, and these fatty acids are usually present in the form of alkanoates.
Unlike proteins and rubber, which are made up of amino acids and polymers of isoprene respectively, lipids are the primary class of biomolecules where these alkanoate compounds can be found in significant amounts.
Tambaya 13 Rahoto
The volume in cm3 of a 0.12 moldm−3 HCl required to completely neutralize a 20cm3 of 0.20 moldm−3 of NaOH is
Bayanin Amsa
To find the volume of HCl that is required to completely neutralize the NaOH solution, we need to use the concept of a neutralization reaction. A neutralization reaction occurs when an acid and a base react to form water and a salt, thus neutralizing each other.
In this particular reaction, the balanced chemical equation is:
HCl + NaOH → NaCl + H2O
Here, the equation tells us that one mole of HCl reacts with one mole of NaOH. Therefore, the molar ratio of HCl to NaOH is 1:1.
First, let's determine the number of moles of NaOH present in 20 cm3 solution:
Number of moles of NaOH = Concentration (mol/dm3) × Volume (dm3)
= 0.20 mol/dm3 × 20 cm3 × (1 dm3 / 1000 cm3)
= 0.20 × 0.020
= 0.004 moles
Since the reaction is in a 1:1 ratio, the number of moles of HCl required is also 0.004 moles.
Now, let's determine the volume of HCl solution required:
Volume of HCl (dm3) = Number of moles / Concentration
= 0.004 moles / 0.12 mol/dm3
= 0.03333 dm3
Convert this volume from dm3 to cm3:
0.03333 dm3 × 1000 cm3 / dm3 = 33.33 cm3
Therefore, the volume of HCl required is 33.33 cm3.
Tambaya 14 Rahoto
Which of the following is an air pollutant?
Bayanin Amsa
An air pollutant is any substance in the air, introduced by natural or human activity, that causes harm or discomfort to living organisms, or damages the environment. Let's analyze the substances mentioned:
1. O2 (Oxygen)
Oxygen is the gas we need to breathe. It's not considered an air pollutant because it is essential for human and animal life, as well as many natural processes.
2. CO (Carbon Monoxide)
Carbon Monoxide is a colorless, odorless gas that is produced by burning fuel (like in cars and factories). This gas can be very dangerous if there is a lot of it, as it can prevent oxygen from entering the bloodstream. Because of its harmful effects, it is considered an air pollutant.
3. H2 (Hydrogen)
Hydrogen, while a flammable gas, is generally not harmful to the air or to organisms when it is released into the environment. Therefore, it is not considered an air pollutant.
4. O3 (Ozone)
Ozone is a bit tricky because it is both good and bad. Higher up in the atmosphere, it forms a layer that protects us from the sun’s UV radiation. However, at ground level, it is a harmful air pollutant. Ground-level ozone can cause health problems such as respiratory difficulties, so in this context, it is considered an air pollutant.
In conclusion, the substances that are considered air pollutants in this context are Carbon Monoxide (CO) and ground-level Ozone (O3).
Tambaya 15 Rahoto
A gas when mixed with oxygen, it produces a very hot and early controllable flame. What is the name of the flame and where is it used?
Bayanin Amsa
The Oxy-ethylene flame is a type of flame produced when oxygen is mixed with a gas called ethylene. This mixture results in a flame that is extremely hot and can be easily controlled. Such a flame is often used in industrial applications related to cutting and welding metals. The heat generated by an oxy-ethylene flame is sufficient to melt metals, allowing them to be welded together or cut apart efficiently.
Tambaya 16 Rahoto
An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].
Bayanin Amsa
To find the molecular formula of the compound, follow these steps:
1. Determine the Empirical Formula:
Start by assuming you have 100 grams of the compound. This means you have:
Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):
Next, divide each by the smallest number of moles to get the simplest ratio:
This gives us the empirical formula: C5H7NO2.
2. Determine the Molecular Formula:
The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.
Calculate the empirical formula mass:
The molar mass can be calculated from the vapor density:
Now, find the ratio of the molar mass to the empirical formula mass:
This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.
Therefore, the molecular formula is C5H7NO2.
Tambaya 17 Rahoto
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Bayanin Amsa
The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH3)3CH2CH=C(CH3)3. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:
1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).
2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.
3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.
4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.
Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.
Tambaya 18 Rahoto
The volume occupied by 1 mole of an ideal gas at a temperature of 130 C and a pressure of 1.58 atm is
[ R = 0.082 atm dm3 K−1 mol−1 ]
Bayanin Amsa
According to the Ideal gas equation, PV = nRT
Given: P = 1.58 atm, V = ?, n = 1 mole, R = 0.082, T= 13 + 273K = 286K
Substituting all the given parameters,
V = nRTP
V = 1×0.082×2861.58
V = 14.84 dm3
Tambaya 19 Rahoto
How many isomers has the organic compound represented by the formula C3 H8 O ?
Bayanin Amsa
The molecular formula C3H8O represents organic compounds that contain 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Let's elucidate the possible isomers, which are molecules with the same molecular formula but different structural arrangements.
1. Alcohols: One class of compounds that can form isomers for this formula are alcohols, which include a functional group -OH.
a. Propan-1-ol: This is a straight-chain alcohol where the -OH group is on the first carbon. The structure is as follows:
CH3-CH2-CH2-OH
b. Propan-2-ol: This is another alcohol where the -OH group is on the second carbon, giving it a different structure and properties:
CH3-CH(OH)-CH3
2. Ethers: This is another class of possible isomers, where the oxygen atom is bonded to two alkyl groups.
c. Methoxyethane: Also known as ethyl methyl ether, it has a structure where the oxygen is in a bridge position between a methyl group and an ethyl group:
CH3-O-CH2-CH3
These are the possible structural isomers for this molecular formula. Therefore, the compound C3H8O has three isomers overall:
Thus, the answer is three distinct isomers.
Tambaya 20 Rahoto
Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+ , II. Zn2+ , III. Cu2+
Bayanin Amsa
Aqueous solution of sodium hydroxide (NaOH) is a versatile reagent in chemistry, often used to test for the presence of metal ions. When sodium hydroxide is added to solutions containing certain metal ions, it forms precipitates that are characteristic of those ions. Here's how it interacts with each of the mentioned ions:
Calcium ions (Ca2+): When NaOH is added to a solution containing calcium ions, a white precipitate of calcium hydroxide (Ca(OH)2) can form. However, the precipitate is only slightly soluble in water, and this reaction is not the most definitive test for calcium ions.
Zinc ions (Zn2+): When sodium hydroxide is added to a solution containing zinc ions, a white gelatinous precipitate of zinc hydroxide (Zn(OH)2) forms. This precipitate is soluble in excess NaOH, leading to a clear, colorless solution. This reaction is used to test for zinc ions.
Copper ions (Cu2+): When NaOH is added to a solution containing copper ions, a pale blue precipitate of copper(II) hydroxide (Cu(OH)2) forms. This precipitate is insoluble even in excess NaOH, and the formation of this blue precipitate is a common test for copper ions.
Therefore, an aqueous solution of sodium hydroxide can be used to test for the presence of all three ions: calcium (Ca2+), zinc (Zn2+), and copper (Cu2+). The reaction and precipitate formation with each ion serve as indicators of their presence. Thus, the correct answer is:
I, II and III.
Tambaya 21 Rahoto
An organic compound with general formula RCOR' is an
Bayanin Amsa
The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.
Here is a simple breakdown of the terms:
Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.
Tambaya 22 Rahoto
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Bayanin Amsa
To determine the empirical formula of an oxide of sulfur containing 60% of oxygen, we have to understand the concept of empirical formulas, which give the simplest whole-number ratio of atoms of each element in a compound.
Step 1: Assume 100g of the compound. In 100g of the compound:
Step 2: Convert masses to moles. Use the molar mass to find moles.
Step 3: Determine the simplest whole-number ratio.
To find the ratio, divide each mole value by the smallest number of moles calculated:
The simplest ratio of S:O is 1:3.
Thus, the empirical formula of the oxide is SO3.
Tambaya 23 Rahoto
How much of 5g of radioactive element whose half life is 50days remains after 200days?
Bayanin Amsa
To determine how much of a radioactive element remains after a certain period, we use the concept of half-life. The half-life of a substance is the time it takes for half of the initial amount of a radioactive element to decay. In this example, the half-life is given as 50 days.
We want to know how much of a 5g sample remains after 200 days. First, calculate how many half-lives occur in 200 days:
Number of half-lives = Total time elapsed / Half-life = 200 days / 50 days = 4 half-lives
Next, we calculate the remaining amount after each half-life period:
After 200 days, 0.31g of the radioactive element remains.
Tambaya 24 Rahoto
Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are
Bayanin Amsa
In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:
When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.
Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.
Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.
Tambaya 25 Rahoto
If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of
Bayanin Amsa
The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.
Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.
In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.
Tambaya 26 Rahoto
In the conductance of aqueous CuSO4 solution, the current carriers are the
Bayanin Amsa
In the conductance of aqueous CuSO4 solution, the current carriers are the hydrated ions.
Here's why:
The other options can be understood as follows:
The correct answer is therefore hydrated ions because they enable the conduction of electricity through the aqueous solution.
Tambaya 27 Rahoto
The Van der waals forces of attraction operates between
Bayanin Amsa
The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:
Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.
Tambaya 28 Rahoto
Cx Hy O + 5O2 → 4CO2 + 4H2 O
Cx Hy O in the equation is
Bayanin Amsa
Cx Hy O + 5O2 → 4CO2 + 4H2 O
On balancing the equation, we should have
X = 4 , y = 8 and O = 2 ⇒ C4 H8 O2
Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.
We finally have C2 H4 O
Tambaya 29 Rahoto
The IUPAC name of the compound above is
Bayanin Amsa
To determine the IUPAC name of a compound, follow these steps:
Hence, by following these steps, if the bromo and methyl groups are both attached to the second carbon (lowest numbering possible), the IUPAC name of the compound is "2-bromo, 2-methyl butane."
Tambaya 30 Rahoto
Sulphur(IV)oxide can be used as a
Bayanin Amsa
Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide is a good refrigerant because it has a high heat of evaporation and can be easily condensed.
Tambaya 31 Rahoto
The element which can combine with oxygen to form an acid anhydride of the form XO2 is
Bayanin Amsa
An Acid anhydride can be defined as a non-metal oxide which forms an acidic solution when reacted with water.
Sulphur is the element that can combine with oxygen to form an acid anhydride of the form XO2 .
An acid oxide is a compound that forms an acid when it reacts with water. Non-metals in groups 4–7 form acidic oxides.
Tambaya 32 Rahoto
The IUPAC nomenclature of the compound above is
Bayanin Amsa
The IUPAC nomenclature of the compound above is 2-methylpropan-2-ol.
Tambaya 33 Rahoto
Which of the following is used in forming slag in the blast furnace for the extraction of iron?
Bayanin Amsa
In the process of extracting iron in a blast furnace, CaCO3, or calcium carbonate, plays a crucial role in forming slag. Here is a simple and comprehensive explanation of how it works:
1. Role of Calcium Carbonate (CaCO3):
Calcium carbonate is commonly used as a flux in the blast furnace. When it is introduced into the furnace, it undergoes a decomposition reaction due to the high temperatures, breaking down into calcium oxide (CaO) and carbon dioxide (CO2).
2. Formation of Slag:
The calcium oxide (CaO) produced then reacts with silicon dioxide (SiO2) present in the iron ore. This reaction forms a liquid slag of calcium silicate. The slag serves two main functions:
Thus, calcium carbonate (CaCO3) is crucial for forming slag by providing the necessary calcium oxide (CaO) that reacts with impurities to form slag during the extraction of iron in a blast furnace.
Tambaya 34 Rahoto
The main constituent of water-glass is
Bayanin Amsa
The main constituent of water-glass is sodium trioxosilicate(IV). Water-glass, also known as liquid glass, is common terminology for a mixture of sodium silicate and water. The primary chemical component in water-glass is sodium silicate, which includes sodium ions (Na+) bonded with silicate ions (SiO44-).
Essentially, when sodium silicate is dissolved in water, it results in a viscous liquid that can be utilized in various applications such as in cements, passive fire protection, textile and lumber processing, and as a sealant. Sodium trioxosilicate(IV) forms a significant part of this mixture as it reacts with other compounds to create a hardened, glass-like structure when it dries. Therefore, when water-glass is mentioned, it is mostly referring to solutions that have sodium trioxosilicate(IV) as their principal compound.
Tambaya 35 Rahoto
Bayanin Amsa
When a strong acid reacts with a strong base, the result is the formation of a neutral salt. This reaction is a part of a chemical process known as neutralization.
Let's break it down further:
During a neutralization reaction, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Meanwhile, the remaining ions (for example, Na⁺ from NaOH and Cl⁻ from HCl) come together to form a compound known as a salt. This salt does not affect the acidity or basicity of the solution, hence it is considered neutral.
Therefore, the salt formed in such a reaction is a neutral salt, which is what is referred to as a normal salt in the options provided.
Tambaya 36 Rahoto
The quantity of electricity required to deposit 180g of Ag from a molten silver trioxonitrate(V) is
[Ag = 108]
Bayanin Amsa
To determine the quantity of electricity required to deposit 180g of Ag (silver) from molten silver trioxonitrate(V), we need to understand the concept of electrolysis. During electrolysis, a metal can be deposited according to Faraday's laws of electrolysis.
The equivalent weight of a substance is calculated by dividing the atomic mass by the valency. For silver (Ag), the atomic mass is given as 108 and the valency of silver in AgNO3 is 1. This makes the equivalent weight of Ag 108 g/equivalent.
According to Faraday's first law of electrolysis:
Mass of substance deposited = (Equivalent weight × Quantity of electricity (in coulombs) ) / Faraday's constant (96500 C/mol)
Let's calculate the number of equivalents of silver deposited:
Number of equivalents of Ag = Mass of Ag / Equivalent weight = 180 g / 108 g/equivalent = 5/3 equivalents
The quantity of electricity required to deposit 1 equivalent of a substance is 1 Faraday (F) = 96500 C.
Therefore, the total quantity of electricity required:
Quantity of electricity = Number of equivalents × Faraday's constant
Quantity of electricity = (5/3 equivalents) × 1 F = 5/3 F = 1.67 F
Therefore, 1.67 Faraday is required to deposit 180g of Ag from a molten silver trioxonitrate(V).
Tambaya 37 Rahoto
Biodegradable pollutants are not safe in water systems because they can cause
Bayanin Amsa
Biodegradable pollutants are substances that can be broken down by natural processes and microorganisms. However, when they are present in water systems, they can lead to several environmental and health issues. One of the main concerns is their potential to cause ill health. Here's why:
When biodegradable pollutants such as organic waste are introduced into water bodies, they are decomposed by bacteria and other microorganisms. This process consumes dissolved oxygen in the water. As the oxygen levels decrease, aquatic life such as fish and plants may suffer or die due to a lack of oxygen, disrupting the entire aquatic ecosystem.
This situation is known as eutrophication, which can lead to the excessive growth of algae, commonly referred to as algal blooms. These blooms often produce toxins that are harmful to both aquatic life and humans. Furthermore, when this polluted water is used for drinking, agriculture, or recreational purposes, it poses serious health risks to humans. These risks may include gastrointestinal infections, neurological disorders, and skin problems.
In addition, as the pollutants decompose, foul smells may be released, which can affect air quality in the vicinity, although the primary concern with biodegradable pollutants in water is related to how they affect water quality and health.
Therefore, it is crucial to properly manage and treat biodegradable pollutants before they enter water systems to prevent these health hazards. Failure to do so can result in significant environmental and health issues.
Tambaya 38 Rahoto
The IUPAC nomenclature of the complex K4 Fe(CN)6 is
Bayanin Amsa
The compound in question is K4[Fe(CN)6]. To name this complex using IUPAC nomenclature, let's break it down into parts:
Next, consider the oxidation state of Fe:
Finally, we consider the oxidation state of the iron. Since calculations show that it is +2, the complex ion is named based on its oxidation state.
Hence, the IUPAC name of this compound is potassium hexacyanoferrate(II).
Tambaya 39 Rahoto
For chemical reaction to be spontaneous, ∆G must be
Bayanin Amsa
In the context of chemical reactions, the spontaneity of a reaction is determined by the Gibbs Free Energy change, represented by the symbol ΔG. A chemical reaction is considered to be spontaneous if it proceeds on its own without needing continuous external input of energy.
For a reaction to be spontaneous, the value of ∆G must be negative. This is based on the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where:
A negative value for ΔG indicates that the process releases energy and will proceed spontaneously. This means the system is moving towards a lower energy and more stable state, naturally favoring the products over the reactants.
In contrast, a positive ΔG indicates that the reaction is non-spontaneous and requires energy input. If ΔG is zero, the system is at equilibrium, meaning there is no net change taking place, but this doesn't indicate spontaneity.
Therefore, in summary, for a reaction to be spontaneous, ∆G must be negative.
Tambaya 40 Rahoto
The difference in molecular mass between an alkene and alkyne with six carbon per mole is
Bayanin Amsa
To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.
Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is CnH2n. Therefore, for 6 carbon atoms, the molecular formula is C6H12.
Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is CnH2n-2. Therefore, for 6 carbon atoms, the molecular formula is C6H10.
Now let's calculate the molecular masses:
Molecular mass of alkene (C6H12):
Molecular mass of alkyne (C6H10):
The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.
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